Action of the fundamental group Suppose that $M$ is a smooth manifold. Is it true that the fundamental group $\pi_1(M)$ always acts on $M$? If so, how this action is defined?
EDIT: Of course I want my action to be nontrivial, say faithful. Moreover, I expect this action to have a geometric flavour.
I never saw such a statement but I wonder if it is this true. Very often in geometry one is interested in equivariant versions of some homology theories, so it seems that some manifolds are investigated together with some group action. Of course this is rather vaque motivation but reasons for asking questions may be very broad.
 A: EDIT: This probably doesn't answer your question. But what you will often find is the following.
It is common that given $M$ one looks at its universal cover $\tilde{M}$. Then the fundamental group of $M$ acts indeed on the fibers over $M$. In this case, the quotient of $\tilde{M}$ by the action of $\Gamma=\pi_1(M)$ is diffeomorphic to $M$. This is somewhat complicated but maybe the following will help you.
Suppose you start with a manifold $M$, a simply connected manifold $\tilde{M}$, i.e. connected with trivial fundamental group, and a covering map $p:\tilde{M}\rightarrow M$. Fix a point $x\in M$ and another on $y\in\tilde{M}$ with $p(y)=x$. Then each continuous closed path $\gamma$ in $M$ starting at $x$ can be lifted in $\tilde{M}$ to a path $\tilde{\gamma}$ starting at $y$, not necessarily closed, to some point $z$ in $p^{-1}(x)$ (the fiber over $x$). Now two homotopic paths $\gamma_1,\gamma_2$ in $M$ will lift to paths in $\tilde{M}$ having the same endpoints and being homotopic in $\tilde{M}$. This define an action of $\Gamma=\pi_1(M)$ on $\tilde{M}$.
I have omitted the technical and necessary conditions for which this works just to give intuition.
A: You might want to look at a simple case like $S^1 = [0, 1] / 0 \sim 1$, whose fundamental group is $\mathbb Z$. It's true that you can define a faithful action of $\mathbb Z$ on $S^1$: send $1$ to the map $\theta \mapsto \theta + a$, where $a$ is some irrational number. (Alternatively and equivalently: define $n\cdot p$ to be $p + na \bmod 1$, where $a$ is some fixed irrational). 
But this faithful action doesn't seem to capture much of the relationship between the fundamental group and the circle itself. 
On the other hand, any action of $\mathbb Z$ on $S^1$ has to look something like this, for the orbit of a point $p$ under the action will generally be some infinite sequence of points of $S^1$, and this sequence should probably not have a limit, so they end up 'scattered' like the points $\{p + na \mid n \in \mathbb Z \}$ in the example above. 
In short: looking at this simple example leads me to think that there's probably not a generally nice action of $\pi_1(M)$ on $M$. 
